 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014/2/a/solution

1 - e^{x} = - x - x^{2} /2 + O (x^{3})

and

1 - cos x = x^{2} /2 + O (x^{4})

. Squaring the first expression gives

x^{2} + O (x^{3})

, so canceling the leading quadratic power yields

x \to 0 lim \frac{( 1 - e ^{x} ) ^{2}}{1 - cos x} = 2.

Both numerator and denominator vanish, but their leading nonzero coefficients have ratio two.

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